【bzoj3239】Discrete Logging BSGS

Description

Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 2 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that

BL == N (mod P)

Input

Read several lines of input, each containing P,B,N separated by a space,

Output

for each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print “no solution”.

The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat’s theorem that states

B(P-1) == 1 (mod P)
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat’s theorem is that for any m

B(-m) == B(P-1-m) (mod P) .

Sample Input

5 2 15 2 25 2 35 2 45 3 15 3 25 3 35 3 45 4 15 4 25 4 35 4 412345701 2 11111111111111121 65537 1111111111

Sample Output

013203120no solutionno solution19584351462803587

HINT

Source

---

裸题。

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<map>#include<cmath>using namespace std;typedef long long LL;LL ksm(LL a,LL b,LL mod){    LL ans = 1;    while(b)    {        if(b & 1) ans = ((ans % mod) * (a % mod)) % mod;        a = ((a % mod) * (a % mod)) % mod;        b >>= 1;    }    return ans;}map<LL,LL> h;LL BSGS(LL a,LL b,LL mod){    a %= mod; b %= mod;    if(a == 0 && b == 0) return 1;    else if(a == 0) return -1;    h.clear();    LL m = ceil(sqrt(mod));    LL ni = ksm(a,mod - m - 1,mod);    LL t = 1;    h[t] = 0;    for(int i = 1;i < m;i ++)    {        t = (t * (a % mod)) % mod;        if(t != 1 && !h[t]) h[t] = i;    }    for(int i = 0;i < m;i ++)    {        LL ans = h[b];        if(b == 1 || ans)        {            return i * m + ans;         }        b = (b * ni) % mod;    }    return -1;}int main(){    LL mod,a,b;    while(~scanf("%lld%lld%lld",&mod,&a,&b))    {        LL ans = BSGS(a,b,mod);        if(ans != -1) printf("%lld\n",ans);        else puts("no solution");    }    return 0;}